A man lifts a luggage of $10 k g$ from the ground and puts it on his head $1.7 m$ above the ground. Calculate work done by him and gravity on the luggage respectively. Take acceleration due to gravity $= 10 m s^{- 2}$ .

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Solution

Given:
Mass, $m = 10 k g$ ,
Acceleration due to gravity, $g = 10 m s^{- 2}$ ,
Displacement, $s = 1.7 m$ .
Force $F$ is calculated as
$F = m \times g = 10 \times 10 = 100 N$
Work done by the man on luggage $= F o r c e \times d i s p l a c e m e n t$
$W = 100 \times 1.7 = 170 J$
Hence, the required work done by the man on the luggage is $170 J$ .
It can be noted that the work is done by the man on the luggage because he lifts it against the gravitational force of the earth.

Now, since the gravitational force acting on the luggage is in the donwward direction which is opposite to the direction of motion of the luggage. Hence, the work done by the gravity will be negative.
Also, the force applied by the gravity on the luggage is equal to the weight of the luggage, i.e W = 10 x 10 = 100 N.

Thus , the work done by the gravitational force = - W x h = - 100 x 1.7 = - 170 N

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A man lifts a luggage of 10kg from the ground and puts it on his head 1.7m above the ground. Calculate work done by him and gravity on the luggage respectively. Take acceleration due to gravity=10ms−2.

A man lifts a luggage of $10 k g$ from the ground and puts it on his head $1.7 m$ above the ground. Calculate work done by him and gravity on the luggage respectively. Take acceleration due to gravity $= 10 m s^{- 2}$ .